Answer –
Number of turns in the circular coil is given by N = 16
Radius of the coil is given by r = 10 cm = 0.1 m
Cross-section of the coil, A is given by –
$A=\pi {{r}^{2}}=\pi {{(0.1)}^{2}}$
Current in the coil is given by I = 0.75 A
Magnetic field strength is given by B = 5.0 × 10–2 T
Frequency of oscillations of the coil is given by v = 2.0 s–1
Therefore, the magnetic moment is given by –
$M=NIA=16\times 0.75\times n\times {{(0.1)}^{2}}$
M = 0.377 JT-1
Frequency is given by the relation given below –
$v=\frac{1}{2\pi }\sqrt{\frac{MB}{I}}$
Where, I is the moment of inertia of the coil
Rearranging the above formula, it becomes –
$I=\frac{MB}{4{{\pi }^{2}}{{v}^{2}}}=\frac{0.377\times 5\times {{10}^{-2}}}{4{{\pi }^{2}}\times {{(2)}^{2}}}$
$I=1.2\times {{10}^{-4}}kg{{m}^{2}}$
Hence, this is moment of inertia of the coil about its axis of rotation.